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bài 1: <=> 3x2+3x-2x2-2x+x+1=0 <=> x2+2x+1=0 <=>(x+1)2=0<=>x=-1
bài 2: =(x-3)2+1
vì (x-3)2>=0 với mọi x nên (x-3)2+1>=1 => GTNN của x2-6x+10 là 1 khi x=3
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
a) x vô nghiệm
b)<=>(x2-3x+3)(x2-2x+3)-2x2=(x-3)(x-1)(x2-x+3)
=>(x-3)(x-1)(x2-x+3)=0
TH1:x-3=0
=>X=3
TH2:x-1=0
=>x=1
TH3:x2-x+3=0
<=>(-1)2-4(1.3)=-11
vì -11<0
=>x=1 hoặc 3
bạn tự tiếp làm đi dễ mà
a, \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)=10\)
\(\Leftrightarrow x^3-6x^2+12x-8-x\left(x^2-1\right)+6x^2-18x=10\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x=10\)
\(\Leftrightarrow-5x=18\)
\(\Leftrightarrow x=\frac{-18}{5}\)
b, \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow\left[x+1-\left(x-1\right)\right]\left[\left(x+1\right)^2+x^2-1\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\frac{-6}{12}=\frac{-1}{2}\)
c,\(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3+27=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
\(\Leftrightarrow x=-4\)
d, \(x^3-6x^2+12x-7=0\)
\(\Leftrightarrow\left(x-2\right)^3+1=0\)
\(\Leftrightarrow\left(x-2\right)^3=-1\)
\(\Leftrightarrow x-2=-1\)
\(\Leftrightarrow x=1\)
a) (x - 2)3 - x (x + 1) (x - 1) + 6x (x - 3) = 10
x3 - 6x2 + 12x - 8 - x (x2 - 1) + 6x2 - 18x = 10
x3 - 6x2 + 12x - 8 - x3 + x + 6x2 - 18x = 10
-5x = 10 + 8
x = -3,6
b) (x + 1)3 - (x - 1)3 - 6 (x - 1)2 = -10
x3 + 3x2 + 3x + 1 - (x3 - 3x2 + 3x - 1) - 6 (x2 - 2x + 1) = -10
x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6 = -10
12x - 4 = -10
12x = -6
x = -0,5
c) x3 + 3x2 + 3x + 28 = 0
(x + 1)3 + 27 = 0
(x + 1)3 = -27
(x + 1)3 = (-3)3
=> x + 1 = -3 => x = -4
d) x3 - 6x2 + 12x - 7 = 0
x3 - 6x2 + 12x - 8 + 1 = 0
(x - 2)3 = -1
(x - 2)3 = (-1)3
=> x - 2 = -1 => x = 1